Question #175036

Monochromatic X-rays of wavelength 1A° incident on a crystal and are

diffracted. The glancing angles for (100), (110) and (111) planes were found to

be 16.13°, 23.13° and 28.76°. Identify the crystal structure and find its lattice

constant. Hence, find the atomic weight of crystal, given the density is 8.96 g/cc

Expert's answer

= 2d sin

1/d^{2} = (h^{2} + k^{2} + l^{2})/a^{2}

a^{2} = ^{2}(h^{2} + k^{2} + l^{2})/4sin^{2}

a^{2} = (1 A°)^{2}(1^{2} + 0^{2} + 0^{2}) / 4 sin^{2}(16.13°) = 3.24 A°^{2}

a = 1.80 A°

b^{2} = (1 A°)^{2}(1^{2} + 1^{2} + 0^{2}) / 4 sin^{2}(23.13°) = 3.24 A°^{2}

b = 1.80 A°

c^{2} = (1 A°)^{2}(1^{2} + 1^{2} + 1^{2}) / 4 sin^{2}(28.76°) = 3.24 A°^{2}

c = 1.80 A°

FCC crystal (a = b = c)

Volume for cubic a^{3} = (1.80 A°)^{3} = 5.832 A°^{3} = 5.832 x 10^{-24} cm^{3}

In a FCC lattice there rer 8 atoms at eight corners and 6 at face centers,

Now each corner contributes to eight cells so per unit cell contribution is 1/8 x 8 =1 atom. Now similarly each face center contributes to two unit cells so contribution per unit cell by six face centers is equal to 1/2 x 6 = 3 atoms. Hence total no. of atoms per unit cell of FCC lattice is n = (1+3) = 4 atoms.

density = n x M_{w} / (V_{c} x N_{a})

M_{w} = density x V_{c} x N_{a} / n = 8.96 g/cm^{3} x 5.832 x 10^{-24} cm^{3} x 6.02 x 10^{23} / 4 = 7.86 g/ mol

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